Mandelbrot image in every language

Question 1

I always used aMandelbrot image as the 'graphical' version of Hello World in any graphical application I got my hands on. Now it's your guys' turn.

Language must be capable of graphical output or drawing charts (saving image files disallowed)
Render a square image or graph. The size at least 128 and at most 640 across*
The fractal coordinates range from approximately -2-2i to 2+2i
The pixels outside of the Mandelbrot set should be colored according to the number of iterations before the magnitude exceeds 2 (excluding* black & white)
Each iteration count must have a unique color*, and neighboring colors should preferably be easily distinguishable by the eye
The other pixels (presumably inside the Mandelbrot set) must be colored either black or white
At least 99 iterations
ASCII art not allowed

^{* unless limited by the platform,e.g. graphical calculator}

Allowed:
Allowed
Disallowed:

(shrunken images)

Winning conditions:

Shortest version (size in bytes) for each language will get a mention in this post, ordered by size.
No answer will ever be 'accepted' with the button.

Leaderboard:

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body {  text-align: left !important;  display: block !important;}#answer-list {  padding: 10px;  width: 290px;  float: left;}#language-list {  padding: 10px;  width: 290px;  float: left;}table thead {  font-weight: bold;}table td {  padding: 5px;}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script><link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/all.css?v=ffb5d0584c5f"><div>  <h2>Shortest Solution by Language</h2>  <table>    <thead>      <tr><td>Language</td><td>User</td><td>Score</td></tr>    </thead>    <tbody>    </tbody>  </table></div><div>  <h2>Leaderboard</h2>  <table>    <thead>      <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr>    </thead>    <tbody>    </tbody>  </table></div><table>  <tbody>    <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>  </tbody></table><table>  <tbody>    <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>  </tbody></table>

Question 2

"Easily distinguished by the eye" is hard to make objective. ...Also, apart from your personal association of the two, the Mandelbrot set has nothing to do with Hello World, so it's best to omit that from the title unless you are deliberately trolling the search engines.

Question 3

I've seen a few people now mentioning that they render the Mandelbrot as a "Hello World". I've done that too, for something like 30 years. The Mandelbrot is the perfect "Hello World" because it shows both that you have pixel access to the display and gives a good feel for compute bound performance on the new platform.

Question 4

Great idea to ask a question which requires a combination of mathematical and aesthetic sensitivities, then impose all the the design decisions in advance.

Question 5

Anyone manages to make one in brainfuck WINS, I'd say :D

Question 6

Not mine and not golfed, butthis code is a marvel of obfuscated python producing a mandelbrot fractal

Question 7

JavaScript + HTML5 (356B)

(Note: lines ending with '//' are added here for some readability)

Performant version (375B):

<body onload='varw,h=w=C.width=C.height=500,X=C.getContext("2d"),I=X.createImageData(w,h),D=I.data, //y=0,f=255,T=setInterval(function(x,i,j,k,l,c,o){for(x=0;x<w;){                     //for(i=x*4/w-2,j=y*4/h-2,k=l=0,c=f;--c&&k*k+l*l<4;)t=k*k-l*l+i,l=2*k*l+j,k=tD[o=(y*w+x++)*4]=(c*=0xc0ffeeee)&fD[++o]=c>>8&fD[++o]=c>>16&fD[++o]=f}X.putImageData(I,0,0)++y-h||clearInterval(T)},0)'><canvas id=C>

Slow version (356B): remove the 'var' and parameters in the inner function so that the global scope is used.

Try it out:http://jsfiddle.net/neuroburn/Bc8Rh/

enter image description here

Question 8

Forgive me if I don't understand your instructions on making the short version.

Question 9

No problem. Removevar w, at the beginning, and changefunction(x,i,j,k,l,c,o) tofunction().

Question 10

Matlab (89 bytes)

[X,Y]=ndgrid(-2:.01:2);C=X+i*Y;Z=C-C;K=Z;for j=1:99,Z=Z.*Z+C;K=K+(abs(Z)<2);end,imagesc(K)

Output -

enter image description here

Doesn't satisfy the requirement that the inner cells must be black orwhite, but that can be satisfied by either (1) usingimshow(K) insteadofimagesc(K) (requires 1 fewer byte but needs the image processingtoolbox) or (2) appendingcolormap hot (requires 12 more bytes).

Ungolfed version -

Z = zeros(N);K = Z;[X,Y]=ndgrid(-2:.01:2);C = X+1i*Y;for j = 1:99  Z = Z.*Z + C;  K(K==0 & abs(Z) > 2) = j;endimagesc(K)

Question 11

Using a library is fine if it's packaged in Matlab by default and any user can guess it's being used from the code or the error messages.

Question 12

Nice, you beat me. I like theC-C in place of my0*e(401). Plus, you're not usingN. And we can get a little shorter using mym+=abs(z)<2 idea in place of yourK(~K&abs(Z)>2)=j.

Question 13

The defaultcolormap jet andcolormap hot are both incorrect though - they only have 64 distinct colors.colormap(hot(101)) doesn't look visually distinguishable to me.colormap([0,0,0;jet(100)]) is maybe acceptable but iffy.

Question 14

Does that work? On Octave,K=K+abs(Z)<2 meansK=((K+abs(Z))<2). (So I was wrong about the one byte estimate to eliminate+=.)

Question 15

The Mandelbrot set rotated by 90 degrees is still the Mandelbrot set.

Question 16

Javascript, 285B

Based off my code and some improvements onMT0's code, I've got this down to 285B in colour:

document.body.appendChild(V=document.createElement('Canvas'));j=(D=(X=V.getContext('2d')).createImageData(Z=V.width=V.height=255,Z)).data;for(x=Z*Z;x--;){k=a=b=c=0;while(a*a+b*b<4&&Z>k++){c=a*a-b*b+4*(x%Z)/Z-3;b=2*a*b+4*x/(Z*Z)-2;a=c;}j[4*x]=99*k%256;j[4*x+3]=Z;}X.putImageData(D,0,0);

in action:http://jsfiddle.net/acLhe/7/

was: Coffeescript, 342B

document.body.appendChild V=document.createElement 'Canvas'N=99Z=V.width=V.height=400P=[]P.push "rgba(0,0,0,"+Math.random()*i/N+')' for i in [N..0]X=V.getContext '2d'for x in [0..Z] for y in [0..Z]  k=a=b=0  [a,b]=[a*a-b*b+4*x/Z-3,2*a*b+4*y/Z-2] while a*a+b*b<4 and N>k++  X.fillStyle=P[k-1]  X.fillRect x,y,1,1

Coffeescript is supposed to be readable :-/ see it in action:http://jsfiddle.net/acLhe/6/

Mandelbrot Coffeescript

Question 17

OP asks for color unless your platform doesn't support color. Looks great, though, and nice concise code. Welcome to PPCG!

Question 18

I start from this size 285B and improve it more in thisanswer

Question 19

QBasic, QuickBasic, QB64 -156 153

SCREEN 13FOR J=0TO 191B=J/48-2FOR I=0TO 191A=I/48-2X=AY=BC=0DOU=X*XV=Y*YY=2*X*Y+BX=U-V+AC=C+1LOOP UNTIL C>247OR U+V>4PSET(I,J),CNEXTNEXT

Standard DOS palette:

enter image description here

Question 20

Surprising that in all of 11 years no one noticed the ways in which this can be optimized. (I left inefficiencies in there on purpose)

Question 21

Python 3, 290 Bytes

from PIL import Imageimport numpy as ns=640;d=n.zeros((s,s,3),dtype=n.uint8)for k in range(s*s):    r=4/s*(k%s)-2;i=2-4/s*(k//s);z=c=complex(r,i)    for i in range(50,9950,100):        if abs(z)>2:            d[k//s,k%s]=[i*5,i*9,i*7];break        z=z*z+cImage.fromarray(d).show()

Question 22

QBasic, 139 bytes

-1 c/oDLosc

SCREEN 13FOR i=0TO 199FOR j=0TO 199x=0y=0FOR c=0TO 255t=y*yc=c-255*(x*x+t>4)y=2*x*y+j/50-2x=x*x-t+i/50-2NEXTPSET(i,j),cNEXT j,i

Output:

It may be necessary to append aINPUT$(1) to prevent the window from closing until a key is pressed.

FreeBASIC, 139 bytes

UsingFreeBASIC'sSCREEN 16, we can generate a more detailed image at the same code length:

SCREEN 16FOR i=0TO 383FOR j=0TO 383x=0y=0FOR c=0TO 255t=y*yc=c-255*(x*x+t>4)y=2*x*y+j/96-2x=x*x-t+i/96-2NEXTPSET(i,j),cNEXT j,i

Compile asfbc -lang qb mandel.bas.

Higher resolutions are also available.

Question 23

Please count newlines as a byte too

Question 24

@MarkJeronimus my mistake, I though I had.

Question 25

Modifyingc to exit theFOR loop is really clever! You can save 2 bytes by usingy*y directly instead of assigning tot, and one more byte by writingNEXT j,i instead ofNEXT twice. (Not sure if that second one works in FreeBASIC.)

Question 26

The assignment tot is necessary, because the old value ofy is used in the computation ofx. Thanks for the multi-NEXT tip ;)

Question 27

Perl + GD, 264

$I=new GD::Image $s=499,$s;Z(0,0,0);Z(map rand 256,1..3)for0..99;for$x(0..$s){for$y(0..$s){for($H=$K=$c=$t=0;$c++<99&&$H*$H+$K*$K<4;){subZ{$I->colorAllocate(@_)}($H,$K)=($H*$H-$K*$K+4*$x/$s-2,2*$H*$K+4*$y/$s-2)}useGD;$I->setPixel($x,$y,$c<99&&$c)}}print $I->png

$Mandelbrot fractal from Perl+GD$

Golfed fromthis code

Question 28

Nominated: most ugly color scheme.

Question 29

SmileBASIC, 126 bytes

X=RNDF()*4-2Y=RNDF()*4-2@NN=N+16I=X+S*S-T*TT=Y+S*T*2S=IIF N<#L&&S*S+T*T<4GOTO@NGPSET X*50+99,Y*50+99,RGB(99XOR N,N,N)EXEC.

Question 30

Nice use of randoms to save on loops

Question 31

Should make it 240x240, but I guess that would be longer.

Question 32

Code golf is about what you can get away with, not what youshould do.

Question 33

Tcl/Tk, 316

322324336348349351352353354355

Now a shorter version using 3 letter #RGB shorthands style color triplets (instead of #RRGGBB triplets), which results in different colors.

And some more golfing.

rename set sgrid [canvas .c -w 640 -he 640].c cr i 320 320 -i [s p [image c photo -w 640 -h 640]]time {incr xs y 0time {incr ys a 0s b 0s n 0while \$n<99 {s A [expr $a*$a-$b*$b+$x[s f *4/639.-2]]if [s b [expr 2*$a*$b+$y$f]]*$b+[s a $A]*$a>4 breakincr n}$p p [format #%03x [expr $n*41]] -t $x $y} 640} 640

$fractal$

Tcl/Tk, 325

331333345357358360361362364365

I think I would win if the criterium was beauty!

rename set sgrid [canvas .c -w 640 -he 640].c cr i 320 320 -i [s p [image c photo -w 640 -h 640]]time {incr xs y 0time {incr ys a 0s b 0s n 0while \$n<99 {s A [expr $a*$a-$b*$b+$x[s f *4/639.-2]]if [s b [expr 2*$a*$b+$y$f]]*$b+[s a $A]*$a>4 breakincr n}$p p [format #%06x [expr $n*16777215/99]] -t $x $y} 640} 640

Presentation:

$fractal$

Question 34

Nice. You can reduce a few characters (down to 380, I think) by addingrename set s at the top and then replacing all theset bys

Question 35

GLSL, 146 bytes

void mainImage( out vec4 o, vec2 C ) {o -= o; vec2 z;for(;o.z++<99.^^length(z)>2.;)z=mat2(z.rg,-z.g,z)*z+C/99.-2.;o=mod(vec4(o.z)/vec4(4,2,6,0),1.19);}

https://www.shadertoy.com/view/sly3RW

Question 36

no unique color for you

Question 37

For some reason I see 95% animated noise and 5% Mandelbrot

Question 38

Changing the final statement too=vec4(mod(o.z/2.,2.),mod(o.z,2.),mod(o.z/4.,2.),0) might make it colorful (my theory at least)

Question 39

You're right :) Thanks, updated it to fit the rules now. But +25 bytes

Question 40

JavaScript (pixel shader), 88 bytes

m=(x,y,A=0,B=0,n=99)=>n--&&A*A+B*B<4?m(x,y,A*A-B*B+x,2*A*B+y,n):`hsl(${n*7} 99%${n/2}%)`

A pixel shader function that takesx andy between[-2, -2] and[2, 2] and returns CSS<color> (HSL).

m=(x,y,A=0,B=0,n=99)=>n--&&A*A+B*B<4?m(x,y,A*A-B*B+x,2*A*B+y,n):`hsl(${n*7} 99%${n/2}%)`const size = 512;const scale = 4;const ratio = scale / size;const canvas = document.getElementById('canvas');canvas.width = canvas.height = size;const ctx = canvas.getContext('2d');let x, y;for (x = 0; x < size; x++) for (y = 0; y < size; y++) {  ctx.fillStyle = m((x - size / 2) * ratio, (y - size / 2) * ratio);  ctx.fillRect(x, y, 1, 1);}

<canvas/>

Ungolfed and Explained

m = (  x,    // Re(c)  y,    // Im(c)  A=0,  // Re(z[i])  B=0,  // Im(z[i])  n=99  // 99 - i) =>  n-- && // if haven't reached max depth  A * A + B * B < 4 ? // and |z| < 2    m( // recurse      x,      y,      A * A - B * B + x, // Re(z[i] ^ 2 + c)      2 * A * B + y,     // Im(z[i] ^ 2 + c)      n    ) : // else return HSL colour    `hsl(${n * 7} 99% ${n / 2}%)` // n * 7 always unique in mod 360

Question 41

AWK, 198 bytes

BEGIN{a=-2;c=d=1;b=-1;w=140;h=40;m=250;x=(c-a)/w;y=(d-b)/h;for(r=0;r<h;r++){for(c=0;c<w;c++){q=a+c*x;s=d-r*y;for(R=I=t=0;R*R+I*I<=5&&t<m;t++){T=R*R-I*I+q;I=2*R*I+s;R=T}printf t==m?"#":" "}print""}}'

To Test:

awk 'BEGIN{a=-2;c=d=1;b=-1;w=140;h=40;m=250;x=(c-a)/w;y=(d-b)/h;for(r=0;r<h;r++){for(c=0;c<w;c++){q=a+c*x;s=d-r*y;for(R=I=t=0;R*R+I*I<=5&&t<m;t++){T=R*R-I*I+q;I=2*R*I+s;R=T}printf t==m?"#":" "}print""}}'

Color Attempt, 268 bytes

BEGIN{a=-2;c=d=1;b=-1;w=140;h=40;m=250;x=(c-a)/w;y=(d-b)/hsplit("\033[31m \033[33m \033[32m \033[34m \033[35m",C)for(r=0;r<h;r++){for(c=0;c<w;c++){q=a+c*x;s=d-r*yfor(R=I=t=0;R*R+I*I<=5&&t<m;t++){T=R*R-I*I+qI=2*R*I+s;R=T}printf t==m?"#":C[1+int(4*t/m)]" "}print""}}

To test:

awk 'BEGIN{a=-2;c=d=1;b=-1;w=140;h=40;m=250;x=(c-a)/w;y=(d-b)/h;split("\033[31m \033[33m \033[32m \033[34m \033[35m",C);for(r=0;r<h;r++){for(c=0;c<w;c++){q=a+c*x;s=d-r*y;for(R=I=t=0;R*R+I*I<=5&&t<m;t++){T=R*R-I*I+q;I=2*R*I+s;R=T}printf t==m?"#":C[1+int(4*t/m)]" "}print""}}'

Question 42

I really like hacker solutions like this. Please provide a full command line and a screenshot of the output.

Question 43

Cheers, didn't realize it was so easy to add photos.

Question 44

Typst, 210 bytes

(specifically Typst v0.14.0)

#(w=>image(format:(encoding:"rgb8",width:w,height:w),for i in(w=range(w).map(y=>y/w*4-2))+w{for j in w{let(a,b,k)=(0,0,99);while 1>k<(4>a*a+b*b){(a,b)=(-b*b+a*a+j,2*a*b+i);k -=1};(k*k).to-bytes(size:3)}}))(640)

Explanation

First we have some boilerplate to get image output:

#(w=>image(format:(encoding:"rgb8",width:w,height:w), {  ...  } ))(640)

Since we specifyencoding:"rgb8", the positional argument is expected to be abytes object (essentially a tightly-packed array of u8s) of length3*w*w (in the format(r,g,b,r,g,b,r,g,b,...)). Typst automatically joins together the results of consecutive expressions, and, like Rust, considersfor loops,if-else blocks, and the like to be expressions. In particular,bytes objects join together by concatenation, so we can construct abytes object of length (for example) 300 with an expression of the following structure:

for i in range(10){    for j in range(10){        bytes((0,0,0)) //all 100 of these are joined together    }}

We use the same structure in the golf, where the expression within the innerfor loop returns a length-3bytes object (as we will see):

for i in(w=range(w).map(y=>y/w*4-2))+w{for j in w{ {  ...  } }}

In the expression(w=range(w).map(y=>y/w*4-2))+w, the left operandw=range(w).map(y=>y/w*4-2) evaluates first, shadowingw (reusing the same variable name means we don't needlet for this assignment) and evaluating tonone (as all assignments do). The right operand is then evaluated, returning the updated value ofw, and the additionnone + w leaves it unchanged (none can be added to anything without changing it). The code is thus equivalent to if we had just created a new block for the assignment:

{w=range(w).map(y=>y/w*4-2);for i in w{for j in w{ {  ...  } }}}

but our version is one byte shorter for having removed the space betweenin andw.

Inside thefor loops, we have:

let(a,b,k)=(0,0,99);while 1>k<(4>a*a+b*b){(a,b)=(-b*b+a*a+j,2*a*b+i);k -=1};{  ...  }

The expression1>k<(4>a*a+b*b) uses the fact thatbools are comparable (andfalse < true is the only true case) to compute the equivalent ofk>=1 and 4>a*a+b*b. Note that the seemingly obvious byte saves of-b*b+a*a+j ->a*a-b*b+j andk -=1 ->k-=1 are in fact impossible, since Typst permits hyphens in identifier names (including at the end) so will interpreta-b andk- as nonexistent variables.

Finally, we return a length-3bytes object representing the pixel colour, based on the iteration count:

(k*k).to-bytes(size:3)

In other words, we get the three least-significant bytes (sinceto-bytes defaults to little-endian) ofk*k. This achieves a fairly high degree of dispersion (particularly when the iteration count is lower, and hencek is larger) for a low byte count, and keeps the points inside the Mandelbrot set (for whichk is 0) pure black.

Typst, 207 bytes

#(w=>image(format:(encoding:"rgba8",width:w,height:w),for i in(w=range(w).map(y=>y/w*4-2))+w{for j in w{let(a,b,k)=(0,0,1);while k>0>(a*a+b*b>4){(a,b)=(-b*b+a*a+j,2*a*b+i);k/=2.1};k.to-bytes(size:4)}}))(640)

Mostly works the same as the previous solution, except that instead of usingk as an integer and counting down, we make it a float and repeatedly divide it by 2.1, with the exit condition being float underflow (k will eventually be so small that it is rounded to 0); we then take the colour from the 4 (little-endian) bytes of a single-precision representation ofk, with the 4th byte representing alpha (resulting in everything being a bit washed out).We lose one byte toencoding:"rgba8", but save 4 bytes in the main body:

let(a,b,k)=(0,0,1);while k>0>(a*a+b*b>4){(a,b)=(-b*b+a*a+j,2*a*b+i);k/=2.1};k.to-bytes(size:4)

This is horribly inefficient, since it takes about 1000 iterations fork to reach 0, but when rounded to single-precision it reaches 0 in only 140 iterations, so we compute far more iterations than we display. Nonetheless, this means we correctly display >99 iterations, meeting the challenge minimum.

Typst, 205 bytes

#(w=>image(format:(encoding:"rgba8",width:w,height:w),for i in(w=range(w).map(y=>y/w*4-2))+w{for j in w{let(a,b,k)=(0,0,1);while k>0>(a*a+b*b>4){(a,b)=(-b*b+a*a+j,2*a*b+i);k/=2};k.to-bytes(size:4)}}))(640)

The previous solution, but with 2.1 cut down to 2. For obvious reasons, this makes the bytes of the floating point representations much more predictable; however, it luckily happens to produce a near-alternating pattern (so the neighbouring colours are clearly distinguishable), and each seemingly-identical colour is actually a very slightly different shade, so the challenge conditions are again met. (The number of displayed iterations is 150 in this case. Note that we can't use a divisor of 3, since that reaches single-precision zero by iteration 95).

Question 45

How cheeky of you to use a float as both the iteration counter and as a color. I learned a new trick.

Question 46

Julia

Well, better late than never:

function mandelbrot(x0,y0,side,N=800,L=55,R=3.)    m = [0 for i=1:N,j=1:N]    delta = side/N    for i=1:N, j=1:N        c = x0+delta*i+(y0+delta*j)*im        z, h = 0+0*im, 0        while (h<L) && (abs(z)<R)            z = z^2+c            h+=1        end        m[j,i]=h    end    return mendn=2.6m = mandelbrot(-n/1.3,-n/2, n)using Winston, Colorimagesc(m)title("Mandelbrot Set")

Colormap Mod. :

function RGB_cm()    colormap = [RGB(0,0,0) for t=1:255*5]    rgb = [255,0,0]    for t in 0:(255*5-1)    c = [0, 0, 0]    i = ifloor(t/255)    c[(i+3)%3!=0?(i+3)%3:3] = (-1)^i    rgb+=c    colormap[t+1] = RGB(rgb[1],rgb[2],rgb[3])    end    colormap[(end-25):end] = RGB(0,0,0)    return colormapendc = RGB_cm()Winston.colormap(c)

Output:

Mandelbrot Set

Question 47

Should I byte-count only the first program or both?

Question 48

I think both, because without the color map mod the image would not satisfy the rules.

Question 49

I'm sure you can save much on whitespace.

Question 50

CASIO BASIC (CASIO fx-9750GIII), 96 bytes

translation of12Me21's Answer it TI-80 BASIC

For -2→R To 2 Step .1For -2→Q To 2 Step .10→S~TFor 1→N To 21SS+TT≥4⇒BreakSS-TT+Q→I2ST+R→TI→SNextFrac (.5N⇒Plot Q,RNextNext

this is unbelievably slow.

check it out online

Question 51

You can get a ""photo"" of it if you let it run in that online emulator

ɲeuroburɳ 4012 silver badges4 bronze badges · Answer 1 · 2014-03-08 21:27:45Z

JavaScript + HTML5 (356B)

(Note: lines ending with '//' are added here for some readability)

Performant version (375B):

<body onload='varw,h=w=C.width=C.height=500,X=C.getContext("2d"),I=X.createImageData(w,h),D=I.data, //y=0,f=255,T=setInterval(function(x,i,j,k,l,c,o){for(x=0;x<w;){                     //for(i=x*4/w-2,j=y*4/h-2,k=l=0,c=f;--c&&k*k+l*l<4;)t=k*k-l*l+i,l=2*k*l+j,k=tD[o=(y*w+x++)*4]=(c*=0xc0ffeeee)&fD[++o]=c>>8&fD[++o]=c>>16&fD[++o]=f}X.putImageData(I,0,0)++y-h||clearInterval(T)},0)'><canvas id=C>

Slow version (356B): remove the 'var' and parameters in the inner function so that the global scope is used.

Try it out:http://jsfiddle.net/neuroburn/Bc8Rh/

enter image description here

Forgive me if I don't understand your instructions on making the short version.
No problem. Removevar w, at the beginning, and changefunction(x,i,j,k,l,c,o) tofunction().

Chris Taylor 1574 bronze badges · Answer 2 · 2014-03-11 13:28:04Z

4

\$\begingroup\$

Matlab (89 bytes)

[X,Y]=ndgrid(-2:.01:2);C=X+i*Y;Z=C-C;K=Z;for j=1:99,Z=Z.*Z+C;K=K+(abs(Z)<2);end,imagesc(K)

Output -

enter image description here

Doesn't satisfy the requirement that the inner cells must be black orwhite, but that can be satisfied by either (1) usingimshow(K) insteadofimagesc(K) (requires 1 fewer byte but needs the image processingtoolbox) or (2) appendingcolormap hot (requires 12 more bytes).

Ungolfed version -

Z = zeros(N);K = Z;[X,Y]=ndgrid(-2:.01:2);C = X+1i*Y;for j = 1:99  Z = Z.*Z + C;  K(K==0 & abs(Z) > 2) = j;endimagesc(K)

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editedMar 11, 2014 at 13:28

answeredMar 10, 2014 at 10:57

Chris Taylor

1574 bronze badges

\$\endgroup\$

7

\$\begingroup\$Using a library is fine if it's packaged in Matlab by default and any user can guess it's being used from the code or the error messages.\$\endgroup\$
Mark Jeronimus
– Mark Jeronimus
2014-03-10 19:19:52 +00:00
CommentedMar 10, 2014 at 19:19
\$\begingroup\$Nice, you beat me. I like theC-C in place of my0*e(401). Plus, you're not usingN. And we can get a little shorter using mym+=abs(z)<2 idea in place of yourK(~K&abs(Z)>2)=j.\$\endgroup\$
aschepler
– aschepler
2014-03-11 12:33:45 +00:00
CommentedMar 11, 2014 at 12:33
\$\begingroup\$The defaultcolormap jet andcolormap hot are both incorrect though - they only have 64 distinct colors.colormap(hot(101)) doesn't look visually distinguishable to me.colormap([0,0,0;jet(100)]) is maybe acceptable but iffy.\$\endgroup\$
aschepler
– aschepler
2014-03-11 12:39:52 +00:00
CommentedMar 11, 2014 at 12:39
\$\begingroup\$Does that work? On Octave,K=K+abs(Z)<2 meansK=((K+abs(Z))<2). (So I was wrong about the one byte estimate to eliminate+=.)\$\endgroup\$
aschepler
– aschepler
2014-03-11 13:06:41 +00:00
CommentedMar 11, 2014 at 13:06
2
\$\begingroup\$The Mandelbrot set rotated by 90 degrees is still the Mandelbrot set.\$\endgroup\$
Chris Taylor
– Chris Taylor
2015-03-05 23:19:23 +00:00
CommentedMar 5, 2015 at 23:19

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alyx-brett 1,5738 silver badges14 bronze badges · Answer 3 · 2020-06-17 09:04:33Z

Javascript, 285B

Based off my code and some improvements onMT0's code, I've got this down to 285B in colour:

document.body.appendChild(V=document.createElement('Canvas'));j=(D=(X=V.getContext('2d')).createImageData(Z=V.width=V.height=255,Z)).data;for(x=Z*Z;x--;){k=a=b=c=0;while(a*a+b*b<4&&Z>k++){c=a*a-b*b+4*(x%Z)/Z-3;b=2*a*b+4*x/(Z*Z)-2;a=c;}j[4*x]=99*k%256;j[4*x+3]=Z;}X.putImageData(D,0,0);

in action:http://jsfiddle.net/acLhe/7/

was: Coffeescript, 342B

document.body.appendChild V=document.createElement 'Canvas'N=99Z=V.width=V.height=400P=[]P.push "rgba(0,0,0,"+Math.random()*i/N+')' for i in [N..0]X=V.getContext '2d'for x in [0..Z] for y in [0..Z]  k=a=b=0  [a,b]=[a*a-b*b+4*x/Z-3,2*a*b+4*y/Z-2] while a*a+b*b<4 and N>k++  X.fillStyle=P[k-1]  X.fillRect x,y,1,1

Coffeescript is supposed to be readable :-/ see it in action:http://jsfiddle.net/acLhe/6/

Mandelbrot Coffeescript

OP asks for color unless your platform doesn't support color. Looks great, though, and nice concise code. Welcome to PPCG!
I start from this size 285B and improve it more in thisanswer

Mark Jeronimus 7,2266 gold badges19 silver badges23 bronze badges · Answer 4 · 2014-03-14 18:35:36Z

4

\$\begingroup\$

QBasic, QuickBasic, QB64 -156 153

SCREEN 13FOR J=0TO 191B=J/48-2FOR I=0TO 191A=I/48-2X=AY=BC=0DOU=X*XV=Y*YY=2*X*Y+BX=U-V+AC=C+1LOOP UNTIL C>247OR U+V>4PSET(I,J),CNEXTNEXT

Standard DOS palette:

enter image description here

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editedMar 14, 2014 at 18:35

answeredMar 14, 2014 at 18:29

Mark Jeronimus

7,2266 gold badges19 silver badges23 bronze badges

\$\endgroup\$

1

\$\begingroup\$Surprising that in all of 11 years no one noticed the ways in which this can be optimized. (I left inefficiencies in there on purpose)\$\endgroup\$
Mark Jeronimus
– Mark Jeronimus
2025-03-15 11:52:12 +00:00
CommentedMar 15 at 11:52

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r3dapple 1414 bronze badges · Answer 5 · 2019-10-06 16:38:58Z

Python 3, 290 Bytes

from PIL import Imageimport numpy as ns=640;d=n.zeros((s,s,3),dtype=n.uint8)for k in range(s*s):    r=4/s*(k%s)-2;i=2-4/s*(k//s);z=c=complex(r,i)    for i in range(50,9950,100):        if abs(z)>2:            d[k//s,k%s]=[i*5,i*9,i*7];break        z=z*z+cImage.fromarray(d).show()

primo 33.7k5 gold badges63 silver badges142 bronze badges · Answer 6 · 2021-06-04 11:55:04Z

QBasic, 139 bytes

-1 c/oDLosc

SCREEN 13FOR i=0TO 199FOR j=0TO 199x=0y=0FOR c=0TO 255t=y*yc=c-255*(x*x+t>4)y=2*x*y+j/50-2x=x*x-t+i/50-2NEXTPSET(i,j),cNEXT j,i

Output:

It may be necessary to append aINPUT$(1) to prevent the window from closing until a key is pressed.

FreeBASIC, 139 bytes

UsingFreeBASIC'sSCREEN 16, we can generate a more detailed image at the same code length:

SCREEN 16FOR i=0TO 383FOR j=0TO 383x=0y=0FOR c=0TO 255t=y*yc=c-255*(x*x+t>4)y=2*x*y+j/96-2x=x*x-t+i/96-2NEXTPSET(i,j),cNEXT j,i

Compile asfbc -lang qb mandel.bas.

Higher resolutions are also available.

Modifyingc to exit theFOR loop is really clever! You can save 2 bytes by usingy*y directly instead of assigning tot, and one more byte by writingNEXT j,i instead ofNEXT twice. (Not sure if that second one works in FreeBASIC.)
The assignment tot is necessary, because the old value ofy is used in the computation ofx. Thanks for the multi-NEXT tip ;)

mob 2,59618 silver badges14 bronze badges · Answer 7 · 2014-03-09 06:24:30Z

Perl + GD, 264

$I=new GD::Image $s=499,$s;Z(0,0,0);Z(map rand 256,1..3)for0..99;for$x(0..$s){for$y(0..$s){for($H=$K=$c=$t=0;$c++<99&&$H*$H+$K*$K<4;){subZ{$I->colorAllocate(@_)}($H,$K)=($H*$H-$K*$K+4*$x/$s-2,2*$H*$K+4*$y/$s-2)}useGD;$I->setPixel($x,$y,$c<99&&$c)}}print $I->png

$Mandelbrot fractal from Perl+GD$

Golfed fromthis code

12Me21 6,7911 gold badge23 silver badges40 bronze badges · Answer 8 · 2020-06-17 09:04:33Z

3

\$\begingroup\$

SmileBASIC, 126 bytes

X=RNDF()*4-2Y=RNDF()*4-2@NN=N+16I=X+S*S-T*TT=Y+S*T*2S=IIF N<#L&&S*S+T*T<4GOTO@NGPSET X*50+99,Y*50+99,RGB(99XOR N,N,N)EXEC.

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editedJun 17, 2020 at 9:04

CommunityBot

1

answeredFeb 15, 2017 at 18:50

12Me21

6,7911 gold badge23 silver badges40 bronze badges

\$\endgroup\$

3

\$\begingroup\$Nice use of randoms to save on loops\$\endgroup\$
Mark Jeronimus
– Mark Jeronimus
2017-02-19 09:07:20 +00:00
CommentedFeb 19, 2017 at 9:07
\$\begingroup\$Should make it 240x240, but I guess that would be longer.\$\endgroup\$
snail_
– snail_
2017-03-20 12:28:34 +00:00
CommentedMar 20, 2017 at 12:28
1
\$\begingroup\$Code golf is about what you can get away with, not what youshould do.\$\endgroup\$
12Me21
– 12Me21
2017-06-17 15:28:03 +00:00
CommentedJun 17, 2017 at 15:28

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sergiol 4,1551 gold badge17 silver badges37 bronze badges · Answer 9 · 2020-06-17 09:04:33Z

Tcl/Tk, 316

322324336348349351352353354355

Now a shorter version using 3 letter #RGB shorthands style color triplets (instead of #RRGGBB triplets), which results in different colors.

And some more golfing.

rename set sgrid [canvas .c -w 640 -he 640].c cr i 320 320 -i [s p [image c photo -w 640 -h 640]]time {incr xs y 0time {incr ys a 0s b 0s n 0while \$n<99 {s A [expr $a*$a-$b*$b+$x[s f *4/639.-2]]if [s b [expr 2*$a*$b+$y$f]]*$b+[s a $A]*$a>4 breakincr n}$p p [format #%03x [expr $n*41]] -t $x $y} 640} 640

$fractal$

Tcl/Tk, 325

331333345357358360361362364365

I think I would win if the criterium was beauty!

rename set sgrid [canvas .c -w 640 -he 640].c cr i 320 320 -i [s p [image c photo -w 640 -h 640]]time {incr xs y 0time {incr ys a 0s b 0s n 0while \$n<99 {s A [expr $a*$a-$b*$b+$x[s f *4/639.-2]]if [s b [expr 2*$a*$b+$y$f]]*$b+[s a $A]*$a>4 breakincr n}$p p [format #%06x [expr $n*16777215/99]] -t $x $y} 640} 640

Presentation:

$fractal$

Nice. You can reduce a few characters (down to 380, I think) by addingrename set s at the top and then replacing all theset bys

Max 1614 bronze badges · Answer 10 · 2021-11-18 14:08:57Z

3

\$\begingroup\$

GLSL, 146 bytes

void mainImage( out vec4 o, vec2 C ) {o -= o; vec2 z;for(;o.z++<99.^^length(z)>2.;)z=mat2(z.rg,-z.g,z)*z+C/99.-2.;o=mod(vec4(o.z)/vec4(4,2,6,0),1.19);}

https://www.shadertoy.com/view/sly3RW

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editedNov 18, 2021 at 14:08

answeredNov 18, 2021 at 11:24

Max

1614 bronze badges

\$\endgroup\$

4

\$\begingroup\$no unique color for you\$\endgroup\$
Max
– Max
2021-11-18 11:28:38 +00:00
CommentedNov 18, 2021 at 11:28
\$\begingroup\$For some reason I see 95% animated noise and 5% Mandelbrot\$\endgroup\$
Mark Jeronimus
– Mark Jeronimus
2021-11-18 13:51:28 +00:00
CommentedNov 18, 2021 at 13:51
\$\begingroup\$Changing the final statement too=vec4(mod(o.z/2.,2.),mod(o.z,2.),mod(o.z/4.,2.),0) might make it colorful (my theory at least)\$\endgroup\$
Mark Jeronimus
– Mark Jeronimus
2021-11-18 13:53:59 +00:00
CommentedNov 18, 2021 at 13:53
\$\begingroup\$You're right :) Thanks, updated it to fit the rules now. But +25 bytes\$\endgroup\$
Max
– Max
2021-11-18 14:10:09 +00:00
CommentedNov 18, 2021 at 14:10

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Matthew Jensen 2,3678 silver badges25 bronze badges · Answer 11 · 2022-04-21 23:26:52Z

JavaScript (pixel shader), 88 bytes

m=(x,y,A=0,B=0,n=99)=>n--&&A*A+B*B<4?m(x,y,A*A-B*B+x,2*A*B+y,n):`hsl(${n*7} 99%${n/2}%)`

A pixel shader function that takesx andy between[-2, -2] and[2, 2] and returns CSS<color> (HSL).

m=(x,y,A=0,B=0,n=99)=>n--&&A*A+B*B<4?m(x,y,A*A-B*B+x,2*A*B+y,n):`hsl(${n*7} 99%${n/2}%)`const size = 512;const scale = 4;const ratio = scale / size;const canvas = document.getElementById('canvas');canvas.width = canvas.height = size;const ctx = canvas.getContext('2d');let x, y;for (x = 0; x < size; x++) for (y = 0; y < size; y++) {  ctx.fillStyle = m((x - size / 2) * ratio, (y - size / 2) * ratio);  ctx.fillRect(x, y, 1, 1);}

<canvas/>

Ungolfed and Explained

m = (  x,    // Re(c)  y,    // Im(c)  A=0,  // Re(z[i])  B=0,  // Im(z[i])  n=99  // 99 - i) =>  n-- && // if haven't reached max depth  A * A + B * B < 4 ? // and |z| < 2    m( // recurse      x,      y,      A * A - B * B + x, // Re(z[i] ^ 2 + c)      2 * A * B + y,     // Im(z[i] ^ 2 + c)      n    ) : // else return HSL colour    `hsl(${n * 7} 99% ${n / 2}%)` // n * 7 always unique in mod 360

xrs 2,0471 silver badge9 bronze badges · Answer 12 · 2024-11-29 15:29:40Z

AWK, 198 bytes

BEGIN{a=-2;c=d=1;b=-1;w=140;h=40;m=250;x=(c-a)/w;y=(d-b)/h;for(r=0;r<h;r++){for(c=0;c<w;c++){q=a+c*x;s=d-r*y;for(R=I=t=0;R*R+I*I<=5&&t<m;t++){T=R*R-I*I+q;I=2*R*I+s;R=T}printf t==m?"#":" "}print""}}'

To Test:

awk 'BEGIN{a=-2;c=d=1;b=-1;w=140;h=40;m=250;x=(c-a)/w;y=(d-b)/h;for(r=0;r<h;r++){for(c=0;c<w;c++){q=a+c*x;s=d-r*y;for(R=I=t=0;R*R+I*I<=5&&t<m;t++){T=R*R-I*I+q;I=2*R*I+s;R=T}printf t==m?"#":" "}print""}}'

Color Attempt, 268 bytes

BEGIN{a=-2;c=d=1;b=-1;w=140;h=40;m=250;x=(c-a)/w;y=(d-b)/hsplit("\033[31m \033[33m \033[32m \033[34m \033[35m",C)for(r=0;r<h;r++){for(c=0;c<w;c++){q=a+c*x;s=d-r*yfor(R=I=t=0;R*R+I*I<=5&&t<m;t++){T=R*R-I*I+qI=2*R*I+s;R=T}printf t==m?"#":C[1+int(4*t/m)]" "}print""}}

To test:

awk 'BEGIN{a=-2;c=d=1;b=-1;w=140;h=40;m=250;x=(c-a)/w;y=(d-b)/h;split("\033[31m \033[33m \033[32m \033[34m \033[35m",C);for(r=0;r<h;r++){for(c=0;c<w;c++){q=a+c*x;s=d-r*y;for(R=I=t=0;R*R+I*I<=5&&t<m;t++){T=R*R-I*I+q;I=2*R*I+s;R=T}printf t==m?"#":C[1+int(4*t/m)]" "}print""}}'

I really like hacker solutions like this. Please provide a full command line and a screenshot of the output.

sungodmoth 312 bronze badges · Answer 13 · 2025-10-29 22:34:07Z

Typst, 210 bytes

(specifically Typst v0.14.0)

#(w=>image(format:(encoding:"rgb8",width:w,height:w),for i in(w=range(w).map(y=>y/w*4-2))+w{for j in w{let(a,b,k)=(0,0,99);while 1>k<(4>a*a+b*b){(a,b)=(-b*b+a*a+j,2*a*b+i);k -=1};(k*k).to-bytes(size:3)}}))(640)

Explanation

First we have some boilerplate to get image output:

#(w=>image(format:(encoding:"rgb8",width:w,height:w), {  ...  } ))(640)

Since we specifyencoding:"rgb8", the positional argument is expected to be abytes object (essentially a tightly-packed array of u8s) of length3*w*w (in the format(r,g,b,r,g,b,r,g,b,...)). Typst automatically joins together the results of consecutive expressions, and, like Rust, considersfor loops,if-else blocks, and the like to be expressions. In particular,bytes objects join together by concatenation, so we can construct abytes object of length (for example) 300 with an expression of the following structure:

for i in range(10){    for j in range(10){        bytes((0,0,0)) //all 100 of these are joined together    }}

We use the same structure in the golf, where the expression within the innerfor loop returns a length-3bytes object (as we will see):

for i in(w=range(w).map(y=>y/w*4-2))+w{for j in w{ {  ...  } }}

In the expression(w=range(w).map(y=>y/w*4-2))+w, the left operandw=range(w).map(y=>y/w*4-2) evaluates first, shadowingw (reusing the same variable name means we don't needlet for this assignment) and evaluating tonone (as all assignments do). The right operand is then evaluated, returning the updated value ofw, and the additionnone + w leaves it unchanged (none can be added to anything without changing it). The code is thus equivalent to if we had just created a new block for the assignment:

{w=range(w).map(y=>y/w*4-2);for i in w{for j in w{ {  ...  } }}}

but our version is one byte shorter for having removed the space betweenin andw.

Inside thefor loops, we have:

let(a,b,k)=(0,0,99);while 1>k<(4>a*a+b*b){(a,b)=(-b*b+a*a+j,2*a*b+i);k -=1};{  ...  }

The expression1>k<(4>a*a+b*b) uses the fact thatbools are comparable (andfalse < true is the only true case) to compute the equivalent ofk>=1 and 4>a*a+b*b. Note that the seemingly obvious byte saves of-b*b+a*a+j ->a*a-b*b+j andk -=1 ->k-=1 are in fact impossible, since Typst permits hyphens in identifier names (including at the end) so will interpreta-b andk- as nonexistent variables.

Finally, we return a length-3bytes object representing the pixel colour, based on the iteration count:

(k*k).to-bytes(size:3)

In other words, we get the three least-significant bytes (sinceto-bytes defaults to little-endian) ofk*k. This achieves a fairly high degree of dispersion (particularly when the iteration count is lower, and hencek is larger) for a low byte count, and keeps the points inside the Mandelbrot set (for whichk is 0) pure black.

Typst, 207 bytes

#(w=>image(format:(encoding:"rgba8",width:w,height:w),for i in(w=range(w).map(y=>y/w*4-2))+w{for j in w{let(a,b,k)=(0,0,1);while k>0>(a*a+b*b>4){(a,b)=(-b*b+a*a+j,2*a*b+i);k/=2.1};k.to-bytes(size:4)}}))(640)

Mostly works the same as the previous solution, except that instead of usingk as an integer and counting down, we make it a float and repeatedly divide it by 2.1, with the exit condition being float underflow (k will eventually be so small that it is rounded to 0); we then take the colour from the 4 (little-endian) bytes of a single-precision representation ofk, with the 4th byte representing alpha (resulting in everything being a bit washed out).We lose one byte toencoding:"rgba8", but save 4 bytes in the main body:

let(a,b,k)=(0,0,1);while k>0>(a*a+b*b>4){(a,b)=(-b*b+a*a+j,2*a*b+i);k/=2.1};k.to-bytes(size:4)

This is horribly inefficient, since it takes about 1000 iterations fork to reach 0, but when rounded to single-precision it reaches 0 in only 140 iterations, so we compute far more iterations than we display. Nonetheless, this means we correctly display >99 iterations, meeting the challenge minimum.

Typst, 205 bytes

#(w=>image(format:(encoding:"rgba8",width:w,height:w),for i in(w=range(w).map(y=>y/w*4-2))+w{for j in w{let(a,b,k)=(0,0,1);while k>0>(a*a+b*b>4){(a,b)=(-b*b+a*a+j,2*a*b+i);k/=2};k.to-bytes(size:4)}}))(640)

The previous solution, but with 2.1 cut down to 2. For obvious reasons, this makes the bytes of the floating point representations much more predictable; however, it luckily happens to produce a near-alternating pattern (so the neighbouring colours are clearly distinguishable), and each seemingly-identical colour is actually a very slightly different shade, so the challenge conditions are again met. (The number of displayed iterations is 150 in this case. Note that we can't use a divisor of 3, since that reaches single-precision zero by iteration 95).

How cheeky of you to use a float as both the iteration counter and as a color. I learned a new trick.

CCP 6625 silver badges6 bronze badges · Answer 14 · 2020-06-17 09:04:33Z

Julia

Well, better late than never:

function mandelbrot(x0,y0,side,N=800,L=55,R=3.)    m = [0 for i=1:N,j=1:N]    delta = side/N    for i=1:N, j=1:N        c = x0+delta*i+(y0+delta*j)*im        z, h = 0+0*im, 0        while (h<L) && (abs(z)<R)            z = z^2+c            h+=1        end        m[j,i]=h    end    return mendn=2.6m = mandelbrot(-n/1.3,-n/2, n)using Winston, Colorimagesc(m)title("Mandelbrot Set")

Colormap Mod. :

function RGB_cm()    colormap = [RGB(0,0,0) for t=1:255*5]    rgb = [255,0,0]    for t in 0:(255*5-1)    c = [0, 0, 0]    i = ifloor(t/255)    c[(i+3)%3!=0?(i+3)%3:3] = (-1)^i    rgb+=c    colormap[t+1] = RGB(rgb[1],rgb[2],rgb[3])    end    colormap[(end-25):end] = RGB(0,0,0)    return colormapendc = RGB_cm()Winston.colormap(c)

Output:

Mandelbrot Set

I think both, because without the color map mod the image would not satisfy the rules.

madeforlosers 2,0181 silver badge11 bronze badges · Answer 15 · 2025-03-17 13:19:12Z

CASIO BASIC (CASIO fx-9750GIII), 96 bytes

translation of12Me21's Answer it TI-80 BASIC

For -2→R To 2 Step .1For -2→Q To 2 Step .10→S~TFor 1→N To 21SS+TT≥4⇒BreakSS-TT+Q→I2ST+R→TI→SNextFrac (.5N⇒Plot Q,RNextNext

this is unbelievably slow.

check it out online

You can get a ""photo"" of it if you let it run in that online emulator

Movatterモバイル変換

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Julia

CASIO BASIC (CASIO fx-9750GIII), 96 bytes

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